single phase transformer (karady)
DESCRIPTION
lecture notes on single phase transformerTRANSCRIPT
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04/20/23 360 Topic 3a. Transformer 1-Phase 1
Lecture NotesEEE 360George G. Karady
TOPIC 3a Transformer (Part 1: Single Phase)
Read Chapter : 4.1-4.6
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04/20/23 360 Topic 3a. Transformer 1-Phase 2
Lecture 8
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04/20/23 360 Topic 3a. Transformer 1-Phase 3
TransformerMagnetic circuit analysis
• The magnetic circuit consists of a laminated iron core and a winding.
• The AC current in the winding generates an AC magnetic flux in the core.
• The magnetic field is calculated by Ampere’s law:
F = I N = H Lc
where: F is the magnetomotive force
N is the number of turns
I is the current
H is the magnetic field
Lc is he magnetic path length
I
I
N
Lc
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04/20/23 360 Topic 3a. Transformer 1-Phase 4
TransformerMagnetic circuit analysis
• Magnetic flux density (Weber/m2 or Teslas)
B = Hwhere:
– is the permeability in H/m. = o r
– o = 4 10-7 H/m free space permeability (air)
– r relative permeability (air r =1, and iron r= 5000-8000)
– The actual value of r is determined from the B - H magnetization curve of the magnetic material.
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04/20/23 360 Topic 3a. Transformer 1-Phase 5
TransformerMagnetic circuit analysis
• The magnetic circuits are built with laminated core.
• The core is made out of silicon iron sheets.
• B - H curves of three magnetic materials are shown in the figure.
• The permeability of the core is the slope of the B - H curve.
• The operating range is below the saturation (knee) of the curve. This region is more or less linear.
B - H Curves. (B in teslas, H in amp/m)
1.5
1.0
0.5
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04/20/23 360 Topic 3a. Transformer 1-Phase 6
TransformerMagnetic circuit analysis
• The sinusoidal AC flux induces voltage in the coil (Faraday’s Law)• This voltage is equal to the supply voltage if the ohmic voltage drop is neglected.• The induced voltage is:
• The equation for the sinusoidal flux is: (t) = m sin( t)
• Therefore, the expression for the induced voltage becomes:
• The rms value of the induced voltage is:
where: V is in volts, f = 60 Hz, and m in Weber
dt
d
dt
dN)t(e
VN
N fmm
24 44.
)t ( cos Ndt
dN)t(e m
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04/20/23 360 Topic 3a. Transformer 1-Phase 7
TransformerMagnetic circuit analysis
• The substitution of the = I(N A / Lc) in the induced voltage equation yields:
where
is the inductance
• The energy in the magnetic field is integral of the incoming electric power:
• The voltage induced in a second coil placed on the core is:
c
2
L
ANL
dt
diL
dt
di
L
ANN
dt
dNe 2,1
c1222
dt
diL
dt
di
L
AN
dt
dNe
c
2
2I
0
IL2
1diiLdt
dt
diLidteiW
AL
NNL
c
2102,1
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04/20/23 360 Topic 3a. Transformer 1-Phase 8
TransformerMagnetic circuit analysis
Numerical example
The coil current is 1A in the magnetic circuit or the enclosed figure. The flux density in the gap is B = 1.3 T. The iron core carbon steel
Determine – the flux, – magnetic field H,– number of turns, – coil inductance,– magnetic energy, – the required supply voltage
a
b
c
d
w
h
V
Igap
h = 8cm, w = 10cm, c = 3cm, b = 4cm
d = 4cm, a = 3cm, gap = 0.1 cm
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04/20/23 360 Topic 3a. Transformer 1-Phase 9
Lecture 9
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04/20/23 360 Topic 3a. Transformer 1-Phase 10
TransformerIdeal Transformer
• The transformer has laminated iron-core and a primary and secondary winding
• The windings resistance and leakage flux are zero.
• The primary winding is supplied by a sinusoidal voltage V1.
• The V1 voltage drives a magnetizing current through the winding. I1m
N1 N2
Laminated iron core
Primarywinding
Secondarywinding
I1m
V1
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04/20/23 360 Topic 3a. Transformer 1-Phase 11
TransformerIdeal Transformer
• Magnetizing current I1m
generates a magnetization flux m
in the iron core.
• The flux changes more or less in
sinusoidal form.
• The relation between the flux
and voltage is :
Flux generation
N1 N2
max1m
11 Nf44.4dt
dNV
m
V1
I1m
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04/20/23 360 Topic 3a. Transformer 1-Phase 12
TransformerIdeal Transformer
• The ac flux links to the secondary winding.
• The flux change (dm /dt)
induces a sinusoidal voltage V2 in the secondary winding.
• The induced voltage is :
• The ratio of the primary and secondary voltages is called the turn ratio:
a = V1/V2 = N1/N2
Voltage generation
N1 N2
max2m
22 Nf44.4dt
dNV
I1m
V1
m
V2
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04/20/23 360 Topic 3a. Transformer 1-Phase 13
Transformer
Ideal Transformer
• A load impedance Z2 is connected to the secondary.
• The secondary voltage V2 drives a load current through Z2 The current is :
I2 = V2 / Z2
• The load current generates a flux 2 that opposes the
magnetization flux m.
Transformer loaded
N1 N2
Load
I2
2
Z2
I1m
V1
m
V2
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04/20/23 360 Topic 3a. Transformer 1-Phase 14
Transformer
Ideal Transformer
• The load flux 2 induces a voltage in the primary winding that opposes the supply voltage.
• The supply voltage is constant. Therefore the reduction of the induced voltage increases the primary current. (I1m+ I1).
• The I1 current generates a flux 1 that balances and equalizes the flux 2 generated by the secondary current I2.
Transformer loaded
N1 N2
Load
I2
2
m + 1
V2V1
I1m+ I1
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04/20/23 360 Topic 3a. Transformer 1-Phase 15
TransformerIdeal Transformer
• The flux equalization produces the following effects:
– The core flux m remains constant and independent from
the load.– The primary magneto-motive
force F1 is equal to the secondary magnetomotive force F2. Therefore:
I1 N1 = I2 N2
This equation assumes that the magnetizing current Im is negligible small
Transformer loaded
N1 N2
Load
I2I1m+ I1
V1 V2
m + 1- 2
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04/20/23 360 Topic 3a. Transformer 1-Phase 16
TransformerIdeal Transformer
• The losses are zero in an ideal transformer. Therefore the input power (VA) is equal to the output power (VA).
I1 V1 = I2 V2
• The voltage and current relations are:
a = V1 / V2 = I2 / I1 or
V2 =V1/ a and I2 = I1 a
• If a transformer increases the voltage, the current decreases and viceversa.
Equivalent Circuit
V1 = E1
E2 =E 1 / a
I1 I2 = aI1
E2 = V2
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04/20/23 360 Topic 3a. Transformer 1-Phase 17
Transformer
Ideal Transformer. Numerical example
A 50 kVA, 2400 V / 120 V ideal single-phase transformer is loaded with 40
kVA, pf= 0.8 lagging.
a) Draw the circuit diagram and the equivalent circuit.
b) Calculate the primary and secondary voltages and currents.
Circuit diagram 120 V 2400 V
40 kVA
pf = 0.8 lagging
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04/20/23 360 Topic 3a. Transformer 1-Phase 18
TransformerActual Transformer
• The actual transformer windings have resistances R1 and R2 , which are removed from the windings and placed in series with them.
• Part of the primary current generated flux will not link the secondary winding. This flux is the primary leakage flux 11.
• Part of the secondary current generated flux will not link the primary winding. This flux is the secondary leakage flux 22.
I1
N1 mN2
V1V2
I2
11
22
R1 R2
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04/20/23 360 Topic 3a. Transformer 1-Phase 19
Actual Transformer
• The flux linking the primary winding is:
• The flux linking the secondary winding is:
• and 22 can be replaced by equivalent inductance L1 and L2
respectively
Transformer
I1
N1
m
N2
V2E1 E2
R2
I2
R1
V1
L1 L2
1
11
N
LIm11m1
1N2L2I
m22m2
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04/20/23 360 Topic 3a. Transformer 1-Phase 20
Actual Transformer
• In a real transformer the iron core permeability is not infinite and the magnetizing current is not negligible. The iron core is represented by a magnetizing reactance Xm
• Hysteresis and eddy currents cause iron losses. These losses are
represented by a resistance Rc which is connected in parallel with Xm
Transformer
V2E1 E2
V1
X1 R1
Rc Xm
I2
R2 X2
I1
I’1
N1N2
m
IcIm
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04/20/23 360 Topic 3a. Transformer 1-Phase 21
Transformer
Actual Transformer. Equivalent circuit
I1
jX1
Ic
V1 E1
R2jX2
I2 V2E2
I’1R1
Im
RcjXm
Ideal Transformer
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04/20/23 360 Topic 3a. Transformer 1-Phase 22
Transformer
Simplification of equivalent circuit
• The equations for an ideal transformer
E1 = a E2 I1 = I2 /a
The division of the two equation result in
• An impedance can be transferred from one side to the other by multiplying by the square of the turns ratio.
2
22
1
1
I
Ea
I
E 2
21 ZaZ
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04/20/23 360 Topic 3a. Transformer 1-Phase 23
Transformer
Simplification of equivalent circuit
The transfer of the impedances from the secondary to the primary results
in :
I1
jX1
Ic
V1
a2 R2ja2 X2I2
V2E1 = a V2
E2I’2= I2 /a
R1
Im
RcjXm
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04/20/23 360 Topic 3a. Transformer 1-Phase 24
Transformer
Application of equivalent circuit
Classroom exercise 1
A 100 kVA, 7.2 kV / 240 V single phase transformer supplies a variable load (0% - 120%) with pf = 0.8. (lagging). The transformer data are:
R1 = 15 ohm, X1 = j 50 ohm, (Primary side)
R2 = 0.03 ohm, X2 = j 0.05 ohm, (Secondary side)
Rm = 5 kohm, Xm= j 15 kohm, (Primary side)
a) Draw the circuit diagram and equivalent circuit.
b) Plot the supply voltage vs load if the load voltage is 235 V.
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04/20/23 360 Topic 3a. Transformer 1-Phase 25
Lecture 10
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04/20/23 360 Topic 3a. Transformer 1-Phase 26
TransformerEquivalent circuit parameters
• The series impedance of a transformer is calculated from a short circuit test.
• The low voltage side (LV) is short-circuited and the high voltage (HV) is supplied by a reduced voltage which drives rated current through the transformer.
• The voltage, current, and input power are measured.
Short- circuit test
Reduced vo ltage
V
A
W
L VH V
A
sh o rt c irc u it
Isc
V sc
P sc
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04/20/23 360 Topic 3a. Transformer 1-Phase 27
TransformerEquivalent circuit parameters
• The short circuit test gives the current Isc, the supply voltage
Vsc and the power loss Psc.
• The equivalent circuit shows that the series impedance can be calculated from this data.
• The series impedance calculation is:
Ze = Vsc / Isc and Re = Psc / Isc2
Therefore, the equivalent reactance is:2e
2ee RZX
RejXe
Isc secIsc
Vsc
HV LV
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04/20/23 360 Topic 3a. Transformer 1-Phase 28
TransformerEquivalent circuit parameters
• The magnetizing impedance of a transformer is calculated from the open-circuit test.
• The high voltage side (HV) is open and the low voltage side (LV) is supplied by the rated voltage, which drives magnetizing current through the transformer.
• The voltages, current and impute power are measured.
Open - circuit test
A
VW
Vo p enc irc u it
Rated vo ltage
H V L V
P oIo
V oVV
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04/20/23 360 Topic 3a. Transformer 1-Phase 29
TransformerEquivalent circuit parameters
• The open-circuit test gives the magnetizing branch current Io, the supply voltage
V0 and the iron loss Pc.
• The equivalent circuit shows that the magnetizing impedance can be calculated from this data.
• If the series impedance is negligible, the magnetizing impedance is:
R c = V02
/ P0
as So = V0 I0 and
then
20
20m PSQ
0
20
m Q
VX
Re
Rc jXm
jXe
HV LV
Io
Vo
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04/20/23 360 Topic 3a. Transformer 1-Phase 30
TransformerCalculation of equivalent circuit parameters
Numerical example
A 100 kVA, 2400 V / 240 V single phase transformer was tested. In the
short-circuit tests, the HV side was shorted and the parameters were
measured at the LV side. In the open-circuit test, the LV side was opened
and the parameters were measured at the HV side. The results of the tests
are :
• Short-circuit test (HV shorted) Vs = 80 volt, Is = 400 amp, PS = 800 watt
• Open-circuit test (LV open) Vo = 2400 volt, Io = 2 amp, Po = 300 watt
a) Draw the simplified equivalent circuit.
b) Calculate the transformer parameters.
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04/20/23 360 Topic 3a. Transformer 1-Phase 31
Transformer
Questions to ponder.
• Why the discovery of the transformer accelerated the development of the use of electricity ?
• Why the iron or magnetization losses are more important than the losses caused by the winding resistance ?
• Why the transformer has to be cooled ? How it is done ?
• What is the connection of the transformer that supplies your house and where is it ?